Molar Mass and Percent Composition Calculator

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calculate empirical and molecular formula

Empirical and molecular formula from percent composition

The empirical formula gives the smallest whole-number ratio between atoms; the molecular formula says how many atoms the molecule really has. To go from one to the other you only need the molar mass, and the Empirical formula tab does both steps.

From data to formula, in four steps

Take 100 g of the substance, so each percentage becomes grams. Convert each mass to moles by dividing by the atomic mass, divide all the moles by the smallest and, if any quotient is not a whole number, multiply all by the smallest factor that makes everything whole. With 40.0% C, 7.0% H and 53.0% O, the moles are 3.33, 6.94 and 3.31; dividing by 3.31 the ratios are 1.0 : 2.1 : 1.0, or CH2O, with a mass of 30.026 g/mol. Ratios like 1.33 call for a factor of 3, 1.5 for 2 and 1.25 for 4; the tool tries factors from 1 to 6 and accepts deviations of up to 0.05 on the first pass and 0.1 on the second.
The molecular formula is the empirical one multiplied by n = M / M_empirical. With M = 60 g/mol, n = 60 / 30.026 = 2 and the formula is C2H4O2, acetic acid. With M = 180, n = 6 and it is C6H12O6, glucose. The same percent composition, 40% C, 6.7% H and 53.3% O, fits formaldehyde (n = 1), acetic acid and glucose, and only the molar mass breaks the tie. Other classic cases: 92% C and 8% H give CH (13.019 g/mol), which for M = 78 becomes C6H6, benzene; 6% H and 94% O give HO (17.007), which for M = 34 becomes H2O2; 30% N and 70% O give NO2 (46.005), which for M = 92 becomes N2O4.
Ionic compounds have only an empirical formula, because they do not form molecules: the iron oxide with 70% Fe and 30% O gives a ratio of 1 : 1.5 and, doubled, Fe2O3. The oxide with 72% Fe and 28% O gives about 1 : 1.33 and, tripled, Fe3O4, magnetite. Data from real analysis carries experimental error, and the tool warns when the percentages add up far from 100 and when the molar mass entered is not a whole multiple of the empirical one. If the ratios do not come close to small whole numbers, the likeliest cause is a typo or a missing element, and the formula should not be forced.

Composition, molar mass and formula

Simple sugar or acetic acid

Input: C 40 | H 7 | O 53 (M = 60)
Expected output: CH2O → C2H4O2

n = 60 / 30.026 = 2.

Same percentages, another mass

Input: C 40 | H 7 | O 53 (M = 180)
Expected output: CH2O → C6H12O6

n = 180 / 30.026 = 6: glucose.

Benzene

Input: C 92 | H 8 (M = 78)
Expected output: CH → C6H6

Ratio 1 : 1; 78 / 13.019 = 6.

Hydrogen peroxide

Input: H 6 | O 94 (M = 34)
Expected output: HO → H2O2

34 / 17.007 = 2.

Dinitrogen tetroxide

Input: N 30 | O 70 (M = 92)
Expected output: NO2 → N2O4

92 / 46.005 = 2.

Iron(III) oxide

Input: Fe 70 | O 30
Expected output: Fe2O3

Ratio 1 : 1.5, doubled; ionic compound, no molecular formula.

Full tool FAQ

How do you calculate the molar mass of a compound?

Add the atomic weights of every atom in the formula. For aluminum sulfate, Al2(SO4)3: 2 Al + 3 S + 12 O = 2 x 26.9815 + 3 x 32.06 + 12 x 15.999 = 342.131 g/mol. A subscript outside parentheses multiplies everything inside them.

What is the difference between molar mass, molecular mass and molecular weight?

Why does the result differ from my textbook?

How do you calculate the molar mass of a hydrate?

How do you calculate percent composition?

How do you convert grams to moles and molecules?

How do you find the empirical formula?

Does the tool accept ions and isotopes?

Where do the atomic weights come from, and do they change?

Why do some elements show an asterisk or a dagger?

Frequently asked questions

What is the difference between empirical, molecular and minimal formula?

Empirical and minimal are synonyms: the smallest whole-number ratio between atoms. The molecular formula gives the real number of atoms in the molecule and is a multiple of the empirical one. For water both are H2O; for peroxide the empirical is HO and the molecular is H2O2.

What do I do if the ratios come out as 1.33 or 1.5?

Multiply all of them by the same factor until they are whole: 1.33 becomes 4 with factor 3, 1.5 becomes 3 with factor 2. Do not round 1.33 to 1, as that gets the formula wrong. The tool shows the factor used.

Can I use grams instead of percentages?

Yes. Choose Mass in grams and enter how many grams of each element the analysis found. The calculation is the same, because percentages are simply grams in 100 g of sample.

Next steps

Contribution of each element to the molar mass
Element	Atoms	Atomic mass	Subtotal (g/mol)	Mass %
C	6	12.011*	72.066	40.00%
H	12	1.008*	12.096	6.71%
O	6	15.999*	95.994	53.28%
Molar mass	180.156	100%

Contribution of each element to the molar mass

Element

Atoms

Atomic mass

Subtotal (g/mol)

Mass %

12.011*

72.066

40.00%

1.008*

12.096

6.71%

15.999*

95.994

53.28%

Molar mass

180.156

100%

From data to formula, in four steps

Take 100 g of the substance, so each percentage becomes grams. Convert each mass to moles by dividing by the atomic mass, divide all the moles by the smallest and, if any quotient is not a whole number, multiply all by the smallest factor that makes everything whole. With 40.0% C, 7.0% H and 53.0% O, the moles are 3.33, 6.94 and 3.31; dividing by 3.31 the ratios are 1.0 : 2.1 : 1.0, or CH2O, with a mass of 30.026 g/mol. Ratios like 1.33 call for a factor of 3, 1.5 for 2 and 1.25 for 4; the tool tries factors from 1 to 6 and accepts deviations of up to 0.05 on the first pass and 0.1 on the second.

The molecular formula is the empirical one multiplied by n = M / M_empirical. With M = 60 g/mol, n = 60 / 30.026 = 2 and the formula is C2H4O2, acetic acid. With M = 180, n = 6 and it is C6H12O6, glucose. The same percent composition, 40% C, 6.7% H and 53.3% O, fits formaldehyde (n = 1), acetic acid and glucose, and only the molar mass breaks the tie. Other classic cases: 92% C and 8% H give CH (13.019 g/mol), which for M = 78 becomes C6H6, benzene; 6% H and 94% O give HO (17.007), which for M = 34 becomes H2O2; 30% N and 70% O give NO2 (46.005), which for M = 92 becomes N2O4.

Ionic compounds have only an empirical formula, because they do not form molecules: the iron oxide with 70% Fe and 30% O gives a ratio of 1 : 1.5 and, doubled, Fe2O3. The oxide with 72% Fe and 28% O gives about 1 : 1.33 and, tripled, Fe3O4, magnetite. Data from real analysis carries experimental error, and the tool warns when the percentages add up far from 100 and when the molar mass entered is not a whole multiple of the empirical one. If the ratios do not come close to small whole numbers, the likeliest cause is a typo or a missing element, and the formula should not be forced.

Composition, molar mass and formula

Simple sugar or acetic acid

Input: C 40 | H 7 | O 53 (M = 60)
Expected output: CH2O → C2H4O2

n = 60 / 30.026 = 2.

Same percentages, another mass

Input: C 40 | H 7 | O 53 (M = 180)
Expected output: CH2O → C6H12O6

n = 180 / 30.026 = 6: glucose.

Benzene

Input: C 92 | H 8 (M = 78)
Expected output: CH → C6H6

Ratio 1 : 1; 78 / 13.019 = 6.

Hydrogen peroxide

Input: H 6 | O 94 (M = 34)
Expected output: HO → H2O2

34 / 17.007 = 2.

Dinitrogen tetroxide

Input: N 30 | O 70 (M = 92)
Expected output: NO2 → N2O4

92 / 46.005 = 2.

Iron(III) oxide

Input: Fe 70 | O 30
Expected output: Fe2O3

Ratio 1 : 1.5, doubled; ionic compound, no molecular formula.

Full tool FAQ

How do you calculate the molar mass of a compound?

What is the difference between molar mass, molecular mass and molecular weight?

Why does the result differ from my textbook?

How do you calculate the molar mass of a hydrate?

How do you calculate percent composition?

How do you convert grams to moles and molecules?

How do you find the empirical formula?

Does the tool accept ions and isotopes?

Where do the atomic weights come from, and do they change?

Why do some elements show an asterisk or a dagger?

Frequently asked questions

What is the difference between empirical, molecular and minimal formula?

What do I do if the ratios come out as 1.33 or 1.5?

Can I use grams instead of percentages?

Yes. Choose Mass in grams and enter how many grams of each element the analysis found. The calculation is the same, because percentages are simply grams in 100 g of sample.